HMMT 二月 2026 · 团队赛 · 第 4 题

HMMT February 2026 — Team Round — Problem 4

[35] A set of rational numbers S is called inclusive if 0 is not an element of S , and for any (not necessarily distinct) elements x , y , and z of S , the number xy + z is also an element of S . Determine, with proof, all rational numbers a for which there exists an inclusive set containing a .

解析

[35] A set of rational numbers S is called inclusive if 0 is not an element of S , and for any (not necessarily distinct) elements x , y , and z of S , the number xy + z is also an element of S . Determine, with proof, all rational numbers a for which there exists an inclusive set containing a . Proposed by: Henrick Rabinovitz 1 Answer: All rational numbers except 0 and − for all positive integers n . n 1 Solution: We first show that for a ∈ { 0 } ∪ {− | n ∈ N } , there is no inclusive set containing a . By n definition, there is not an inclusive set containing 0 . 1 Let a = − for n ∈ N . Suppose there is an inclusive set S containing a . n k − n Claim 1. For any integer k ≥ 0 , ∈ S . 2 n − n k − n Proof. We use induction on k . The base case k = 0 is true as a = . Now assume ∈ S for a 2 2 n n fixed k . Then ( ) ( ) k − n 1 1 k + 1 − n

− − = ∈ S 2 2 n n n n which completes the inductive step. Plugging in k = n yields that 0 ∈ S , which is a contradiction. Now we show that for all other elements a ∈ Q , there exists an inclusive set containing a . Observe that { x ∈ Q | x > 0 } is an inclusive set as if x, y, z are positive rational numbers, xy + z is as well. m Therefore it suffices to show that if a = − , where m, n ∈ N , m > 1 , and gcd ( m, n ) = 1 , then there n exists an inclusive set containing a . The following claim finishes the problem. Claim 2. The set 2 T = { x ∈ Q | x ≡ a (mod m ) } is an inclusive set containing a . Two rational numbers are equivalent modulo an integer n if their difference in simplest terms has numerator divisible by n . ©2026 HMMT 2 2 Proof. Clearly a ∈ T . Because m ̸ = 1 , m ∤ m , so a ̸ ≡ 0 (mod m ) . Therefore 0 ∈ / T . Lastly, if x , y , and z are in T , then ( ) 2 m 2 2 xy + z ≡ a + a ≡ + a ≡ a (mod m ) n so xy + z ∈ T . As we have checked all necessary conditions, the claim is proven.